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Mean and variance of pdf 12x²(1−x)

Short answer

For f(x)=12x²(1-x) on [0,1], mean E(X)=∫₀¹ x·12(x²−x³)dx=0.6; E(X²)=0.4; variance Var(X)=0.4−0.6²=0.04.

Find the mean and variance of the probability density function given by f(x) = 12x²(1 − x); 0 ≤ x ≤ 1 0 ; otherwise

f(x) = 12x²(1 − x) = 12(x² − x³), 0 ≤ x ≤ 1.

Check: ∫₀¹ 12(x² − x³) dx = 12(1/3 − 1/4) = 12 × 1/12 = 1, so it is a valid pdf.

Mean

E(X) = ∫₀¹ x · 12(x² − x³) dx = 12 ∫₀¹ (x³ − x⁴) dx = 12 (1/4 − 1/5) = 12 × 1/20 = 0.6

Variance

E(X²) = ∫₀¹ x² · 12(x² − x³) dx = 12 ∫₀¹ (x⁴ − x⁵) dx = 12 (1/5 − 1/6) = 12 × 1/30 = 0.4

Var(X) = E(X²) − [E(X)]² = 0.4 − 0.36 = 0.04

Answer: mean = 0.6, variance = 0.04 (standard deviation 0.2)

✓ Answer checked by Claude Sonnet 5.5

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